Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {
	
	public static void main(String[] args) throws IOException {	
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		double a[]=new double[n];
		for(int i=0;i<n;i++)
			a[i]=s.nextDouble();
		double dp[][]=new double[n][n];
		for(int i=0;i<n;i++)
			for(int j=0;j<n;j++)
				dp[i][j]=-1.0;
		System.out.println(String.format("%.10f", rec(0,0,a,n,dp)));
		
	}

	private static double rec(int i, int h, double[] a, int n, double[][] dp) {
		
		if(i==n) {
			if(h>=(n+1)/2)
				return 1.0;
			else
				return 0.0;
		}
		if(dp[i][h]>-1) return dp[i][h];
		double op1=a[i]*rec(i+1, h+1, a, n, dp);
		double op2=(1-a[i])*rec(i+1, h, a, n, dp);
		
		
		return dp[i][h]=op1+op2;
		
	}
	
}